A company wants to test the consistency of its deliveries A

A company wants to test the consistency of its deliveries. A random sample of 25 deliveries yields a sample standard deviation of 2.1 minutes. Specifically, it wishes to test the hypothesis that the population variance is less than or equal to 4. Using an alpha of 0.05, perform the appropriate ChiSquare test on the hypothesis stated above using the critical-value approach clearly showing: Null/Alternate hypotheses, Test statistic, Decision rule, and Conclusion.

Solution

Let o^2 be the variance

The test hypothesis is

Ho: o^2=4 (i.e. null hypothesis)

Ha: o^2 < 4 (i.e. alternative hypothesis)

The test statistic is

chisquare = (n-1)*s^2/o^2

=24*2.1^2/4

=26.46

It is a left-tailed test.

The degree of freedom =n-1=25-1=24

Given a=0.05, the critical value of chisquae with 0.05 and df=24 is 13.85 (from chisqurare table)

The rejection region is if chisquare <13.85, we reject Ho.

Since 26.46 is larger than 13.85, we do not reject Ho.

So we can not conclude that the population variance is less than or equal to 4